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# Evaluate : (i) $$sin 60 ° cos 30° + sin 30° cos 60°$$ (ii) $$2tan^245° + cos^230° - sin^260°$$ (iii) $$\frac{cos45°}{sec30° + cosec30°}$$ (iv) $$\frac{sin30° + tan45°- cosec60°}{sec30° +cos60° +cot45°}$$ (v) $$\frac{5cos^260° +4sec^230° - tan^245°}{sin^230° + cos^230°}$$

(i) $$sin 60° cos 30° \ + \ sin 30° cos 60 \$$
$$= \ \frac{ \sqrt{3}}{2} × \frac{ \sqrt{3}}{2} \ + \ \frac{1}{2} × \frac{1}{2} \$$
$$= \ \frac{3}{4} \ + \ \frac{1}{4} \$$
$$= \ \frac{3+1}{4} \ = \ 1$$

(ii)$$2tan^245° \ + \ cos^230° \ - \ sin^260° \$$
$$= \ 2(1)^2 \ + \ ( \frac{ \sqrt{3}}{2})^2 \ - \ ( \frac{ \sqrt{3}}{2})^2 \$$
$$= \ 2 \ + \ \frac{3}{4} \ - \ \frac{3}{4} \ = \ 2$$

(iii) $$\frac{cos45°}{sec30° + cosec30°} \$$
$$= \ \frac{ \frac{1}{ \sqrt{2}}}{ \frac{2}{ \sqrt{3}} + 2} \$$
$$= \ \frac{ \frac{1}{ \sqrt{2}}}{ \frac{2 + 2 \sqrt{3}}{ \sqrt{3}}} \$$
$$= \ \frac{1}{ \sqrt{2}} × \frac{ \sqrt{3}}{2+2 \sqrt{3}} \$$
$$= \ \frac{ \sqrt{3}}{ \sqrt{2} × 2( \sqrt{3} + 1)} × \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} \$$
$$= \ \frac{ \sqrt{3}( \sqrt{3} - 1)}{ \sqrt{2} × 2(3 - 1)} \$$
$$= \ \frac{ \sqrt{2} × \sqrt{3}( \sqrt{3} - 1)}{ \sqrt{2} × \sqrt{2} × 2 × 2} \$$
$$= \ \frac{3 \sqrt{2} - \sqrt{6}}{8}$$

(iv) $$\frac{sin30° + tan45°- cosec60°}{sec30° +cos60° +cot45°}$$
$$= \ \frac{ \frac{1}{2} + 1 - \frac{2}{ \sqrt{3}}}{ \frac{2}{ \sqrt{3}} + \frac{1}{2} + 1}$$
$$= \ \frac{ \frac{1 + 2}{2} - \frac{2}{ \sqrt{3}}}{ \frac{2}{ \sqrt{3}} + \frac{1 + 2}{2}}$$
$$= \ \frac{ \frac{3}{2} - \frac{2}{ \sqrt{3}}}{ \frac{2}{ \sqrt{3}} + \frac{3}{2}}$$
$$= \ \frac{ \frac{3 \sqrt{3} - 4}{2 \sqrt{3}}}{ \frac{4 + 3 \sqrt{3}}{2 \sqrt{3}}}$$
$$= \ \frac{3 \sqrt{3} - 4}{4 +3 \sqrt{3}}$$
$$= \ \frac{(3 \sqrt{3} - 4)(3 \sqrt{3} - 4)}{(4 + 3 \sqrt{3})(3 \sqrt{3} - 4)}$$
$$= \ \frac{27 + 16 - 12 \sqrt{3} - 12 \sqrt{3}}{27- 16}$$
$$= \ \frac{43 - 24 \sqrt{3}}{11}$$

(v) $$\frac{5cos^260° +4sec^230° - tan^245°}{sin^230° + cos^230°}$$
$$= \ \frac{5( \frac{1}{2})^2 + 4( \frac{2}{ \sqrt{3}})^2 - (1)^2}{( \frac{1}{2})^2 + ( \frac{ \sqrt{3}}{2})^2}$$
$$= \ \frac{5 × \frac{1}{4} + 4 × \frac{4}{3} - 1}{ \frac{1}{4} + \frac{3}{4}}$$
$$= \ \frac{5}{4} \ + \ \frac{16}{3} - 1$$
$$= \ \frac{15 + 64 - 12}{12} \ = \ \frac{67}{12}$$